Re: [CAD_CAM_EDM_DRO] Re: Declining motor torque with lower voltage.

With all the discussion about servo power. Someone needs to and I may when I get time. Do some testing and just find out how a few different sized motors react under different circumstances. Simple sheave mouted on shaft with a cable that lifts a weight. A variac to vary the DC voltage. I'm guessing for X and Y on a Bridgeport that even a nema 23 sized motor of considerable power would suffice. With the proper ratio any servo would work. It just boils down to the rapids that will be achieved. It just doesnt require all that much power for X and Y. In my case, I'm using 40 lb-in motors on X and Y with a 2.5:1 ratio and they are way overkill. I can run roughers all day in steel and the motors never even get warm. Which shows they are doing nothing more than basically freewheeling. Driving the knee for Z on the other hand is a totally different situation. Drilling holes all day, the 40 in-lb motor with 2.5:1 on the knee reduction gets warm, not hot but very warm. So, I conclude using a
2.5:1 ratio driving the knee that a 40in-lb motor is a perfect size with 60 ipm rapids without gas springs or counterweights. A smaller servo may work for this but it will generate more heating and may cause problems and only be suitable for short periods of service. Unless the reduction were increased proportional to the smaller motors size with slower rapids.

Ron

Lance Hopper <snaggletto@...> wrote:


Hey,
Mr. Elson knows his stuff, listen to him. However, I would like
to add, peak torque values mean almost squat. Your 600 oz-in peak
motor probably puts out around 150 oz-in "continuous" torque. This
is the magic number. Furthermore, don't ask your motors run up
to "max RPM" either. Use around 75-80% of max RPM in your rapid
calculations. Good luck.
\
--- In CAD_CAM_EDM_DRO@yahoogroups.com, "Polaraligned"

wrote:

>
>
> > If you need 1000 Pounds of linear force (cutting force + friction

+

> > acceleration)
> > then 1000 *0.0318 = 31.8 In-Lb, or 508 In-Oz. That would

presumably be

> > the PEAK torque requirement for X and Y on a mill. But, it could

be

> > close to this
> > as the CONTINUOUS torque on the knee, unless you had a

counterbalance

> > system.
> >
> > In the above examlple, if you had a 250 Oz-In peak torque motor,

then

> > you'd want a
> > 2:1 belt reduction ratio, at the minimum.
> >
> > Jon
>
> So Jon, what your are saying is that I could direct couple my 600
> oz-in (peak) servo's and they would produce 1000+ lbs of linear

force

> on my Bridgeport. And 1000 lbs is a good value to be at for the
> X and Y axis. I would also then get 240 ipm rapids.
> Am I correct? Everyone seems to be using a lot more power to the
> table than this.
>
> Scott

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